1.) which of the following represents dy/dx when y=e^-2x Sec(3x)?
A.3e^-2x sec(3x) tan (3x)-2e^-2x Sec(3x)<<<< my choice
B.)3e^-2x sec(3x)tan (3x)-2xe^-2x sec(3x)
C.)3e^-2x sec(3x)tan (x)-2e^-2x Sec(3x)
D.)3e^-2x sec(3x)tan (x)-2xe^-2x sec(3x)
2.)calculate dy/dx if y=Ln(2x^3+3x)
A.)1/2x^3+3x
B.)1/6x^2+3
C.)2x^3+3x/6x^2+3
D.)6x^2+3/2x^3+3x <<<< my choice
3.)Given the curve that is described by the equation r=3 cos theta, find the angle that tangent lines makes with the radius vector when theta=120.
A.) 30deg
B.) 45deg
C.) 60deg
D.) 90deg << my choice
4.)A line rotates in a horizontal plane according to the equation theta=2^t-6t,where theta is the angular position of the rotating line, in radians ,and tis the time,in seconds. Determine the angular acceleration when t=2sec.
A.) 6 radians per sec^2
B.) 12 radians per sec^2
C.) 18 radians per sec^2
D.) 24 radians per sec^2 << my choice.
1. Correct
2. Correct
3.
dy/dx
=(dy/dt) / (dx/dt)
put
x=rcos(t)=3cos²(t)=
y=rsin(t)=3sin(t)cos(t)
Calculate dy/dx and evaluate at t=120° (2π/3) to get
dy/dx=-1/√3
=> θ=150°
Angle with radius vector=150°-120°=30°.
4.
There may have been a typo in the question. If the question is asking for θ=2^t-6t, the answers should not be in round numbers.
A
line
rotates
in
a horizontal
plane
according
to
the
equation
9
=
2t3-
6t,
where
(J
is
the
angular
position
of
the
rotating
line,
in
radians,
and
t
is the
time,
in
sec-
onds.
Determine
the
angular
acceleration
when
t
=
2
sec.
1.) The correct choice is B. The derivative of y=e^-2x Sec(3x) with respect to x can be found using the product rule of differentiation.
Let u = e^-2x and v = Sec(3x).
Using the product rule,
dy/dx = u(dv/dx) + v(du/dx).
Now, we need to calculate dv/dx and du/dx.
dv/dx = d(Sec(3x))/dx = 3Sec(3x)tan(3x).
du/dx = d(e^-2x)/dx = -2e^-2x.
Substituting these values back into the formula for dy/dx, we have:
dy/dx = e^-2x (3Sec(3x)tan(3x)) + Sec(3x)(-2e^-2x)
Simplifying, we get:
dy/dx = -2e^-2x Sec(3x) + 3e^-2x Sec(3x)tan(3x)
Therefore, the correct answer is B.
2.) The correct choice is C. To find dy/dx of y = Ln(2x^3+3x), we need to use the chain rule.
Let u = 2x^3 + 3x.
dy/du = 1/u.
du/dx = 6x^2 + 3.
So, using the chain rule:
dy/dx = dy/du * du/dx.
dy/dx = 1/u * (6x^2 + 3).
dy/dx = (6x^2 + 3)/(2x^3 + 3x).
Therefore, the correct answer is C.
3.) The correct choice is C. The equation r = 3 cos theta represents a circle centered at the origin with radius 3. When we differentiate r with respect to theta, we get the slope of the tangent line to the circle at any point.
Differentiating r = 3 cos theta with respect to theta, we get:
dr/dtheta = 3(-sin theta) = -3sin theta.
The slope of the radius vector at theta = 120 degrees is given by -3sin 120 degrees = -3(√3/2) = -3√3/2.
The tangent of the angle made by the tangent line with the radius vector is equal to the negative of the slope, so:
tan angle = -(-3√3/2) = 3√3/2.
The angle can be found by taking the inverse tangent (arctan) of 3√3/2.
angle = arctan(3√3/2) ≈ 60 degrees.
Therefore, the correct answer is C.
4.) The correct choice is B. The angular acceleration is the second derivative of the angular position with respect to time. To find the angular acceleration when t = 2 seconds, we need to take the second derivative of theta with respect to t and plug in t = 2.
Given theta = 2^t - 6t, the first derivative of theta with respect to t is:
d(theta)/dt = 2^(t)ln(2) - 6.
Taking the second derivative, we get:
d^2(theta)/dt^2 = (2^tln(2))^2 - 6.
Evaluating this expression at t = 2:
d^2(theta)/dt^2 = (2^2ln(2))^2 - 6.
d^2(theta)/dt^2 = (4ln(2))^2 - 6.
d^2(theta)/dt^2 = 16ln^2(2) - 6.
Using ln(2) ≈ 0.693, we can calculate:
d^2(theta)/dt^2 ≈ 16(0.693)^2 - 6.
d^2(theta)/dt^2 ≈ 16(0.480) - 6.
d^2(theta)/dt^2 ≈ 7.68 - 6.
d^2(theta)/dt^2 ≈ 1.68.
Therefore, the correct answer is B.
1.) To find dy/dx, we need to apply the product rule and chain rule. Let's break down the expression y=e^-2x Sec(3x):
dy/dx = (d(e^-2x)/dx) * Sec(3x) + e^-2x * (d(Sec(3x))/dx)
Now, let's find the derivatives of e^-2x and Sec(3x):
d(e^-2x)/dx = -2e^-2x (chain rule: derivative of e^-u is -u * e^-u, where u = 2x)
d(Sec(3x))/dx = 3Sec(3x) * Tan(3x) (chain rule: derivative of Sec(u) is Sec(u) * Tan(u), where u = 3x)
Plugging the derivatives back into the equation:
dy/dx = (-2e^-2x) * Sec(3x) + e^-2x * (3Sec(3x) * Tan(3x))
Simplifying:
dy/dx = -2e^-2x Sec(3x) + 3e^-2x Sec(3x) Tan(3x)
Therefore, the correct answer is A. 3e^-2x Sec(3x) Tan(3x) - 2e^-2x Sec(3x).
2.) To find dy/dx, we need to apply the chain rule. Let's break down the expression y=Ln(2x^3+3x):
y = Ln(u) where u = 2x^3 + 3x
dy/dx = (1/u) * du/dx (chain rule: derivative of Ln(u) is (1/u) * du/dx)
Now, let's find the derivative of u = 2x^3 + 3x:
du/dx = 6x^2 + 3 (derivative of 2x^3 is 6x^2, derivative of 3x is 3)
Plugging the derivatives back into the equation:
dy/dx = (1/(2x^3+3x)) * (6x^2 + 3)
Simplifying:
dy/dx = (6x^2 + 3)/(2x^3 + 3x)
Therefore, the correct answer is C. (2x^3 + 3x)/(6x^2 + 3).
3.) To find the angle that the tangent line makes with the radius vector, we need to find the derivative of r with respect to theta, and then use the equation tan(theta) = dy/dx.
Given r = 3cos(theta), let's find dr/dtheta:
dr/dtheta = -3sin(theta) (derivative of cos(theta) is -sin(theta))
Now, let's find dy/dx using the equation tan(theta) = dy/dx:
tan(theta) = dy/dx
tan(theta) = (dy/dtheta)/(dx/dtheta)
tan(theta) = (r * dy/dtheta)/(dr/dtheta) (using polar coordinates: dy = r * dy/dtheta and dx = dr/dtheta)
Substituting values and simplifying:
tan(theta) = (3cos(theta) * dy/dtheta)/(-3sin(theta))
tan(theta) = -cos(theta) * dy/dtheta/sin(theta)
Since the question asks for the angle, we need to find the value of theta at theta = 120 degrees:
tan(120 degrees) = -cos(120 degrees) * dy/dtheta/sin(120 degrees)
Simplifying:
tan(120 degrees) = -(-1/2) * dy/dtheta/(-√3/2)
tan(120 degrees) = (1/2) * (√3/√3)
tan(120 degrees) = √3
Therefore, the correct answer is C. 60 degrees.
4.) To find the angular acceleration when t = 2 seconds, we need to find the second derivative of theta with respect to t. Let's first find the first derivative of theta:
d(theta)/dt = 2^(t) * ln(2) - 6 (using the power rule: derivative of 2^t is (2^t) * ln(2))
Now, let's find the second derivative of theta:
d^2(theta)/dt^2 = (d/dt)(2^(t) * ln(2) - 6)
d^2(theta)/dt^2 = 2^(t) * ln^2(2) (using the power rule: derivative of ln(2) is 0, and derivative of 2^t is (2^t) * ln(2))
Now, let's find the angular acceleration when t = 2 seconds:
d^2(theta)/dt^2 = 2^(2) * ln^2(2)
d^2(theta)/dt^2 = 4 * ln^2(2)
Therefore, the correct answer is B. 12 radians per sec^2.